The reaction $$\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g)$$ has an equilibrium constant of \(1.30\) at \(650^{\circ} \mathrm{C}\). Carbon monoxide and steam both have initial partial pressures of \(0.485 \mathrm{~atm}\), while hydrogen and carbon dioxide start with partial pressures of \(0.159\) atm. (a) Calculate the partial pressure of each gas at equilibrium. (b) Compare the total pressure initially with the total pressure at equilibrium. Would that relation be true of all gaseous systems?

(First, we need to write the equilibrium expression for the given reaction. Based on the stoichiometry, the equilibrium constant (K) expression will be: $$K = \frac{P_{\mathrm{H_{2}}} P_{\mathrm{CO_{2}}}}{P_{\mathrm{CO}} P_{\mathrm{H_{2}O}}}$$)

(To find the partial pressures at equilibrium, we should define the change in partial pressures for each species during the reaction. Let's denote the change in partial pressure as "x". We can represent the changes as follows: $$P_{\mathrm{CO}} = 0.485 - x$$ $$P_{\mathrm{H_{2}O}} = 0.485 - x$$ $$P_{\mathrm{H_{2}}} = 0.159 + x$$ $$P_{\mathrm{CO_{2}}} = 0.159 + x$$)

(Now, we can substitute the expressions for each partial pressure into the equilibrium constant expression: $$1.30 = \frac{(0.159 + x)(0.159 + x)}{(0.485 - x)(0.485 - x)}$$)

(Now, we should solve the equation for "x". This quadratic equation can be simplified as: $$1.30 (0.485 - x)^{2} = (0.159 + x)^{2}$$ Solve for x to find the either of the possible equilibrium values: $$x = 0.258$$ or $$x = -0.602$$ Since negative partial pressures are not possible, we should take the positive value for "x" as: $$x = 0.258$$)

(With the value of "x" determined, we can now calculate the partial pressure of each species at equilibrium: $$P_{\mathrm{CO}} = 0.485 - 0.258 = 0.227 ~\mathrm{atm}$$ $$P_{\mathrm{H_{2}O}} = 0.485 - 0.258 = 0.227 ~\mathrm{atm}$$ $$P_{\mathrm{H_{2}}} = 0.159 + 0.258 = 0.417 ~\mathrm{atm}$$ $$P_{\mathrm{CO_{2}}} = 0.159 + 0.258 = 0.417 ~\mathrm{atm}$$)

(Now, let's compare the total pressure initially and at equilibrium: Initial total pressure: $$P_{\mathrm{total}}^{initial} = P_{\mathrm{CO}}^{initial} + P_{\mathrm{H_{2}O}}^{initial} + P_{\mathrm{H_{2}}}^{initial} + P_{\mathrm{CO_{2}}}^{initial} = 0.485 + 0.485 + 0.159 + 0.159 = 1.288~\mathrm{atm}$$ Final total pressure: $$P_{\mathrm{total}}^{final} = P_{\mathrm{CO}}^{final} + P_{\mathrm{H_{2}O}}^{final} + P_{\mathrm{H_{2}}}^{final} + P_{\mathrm{CO_{2}}}^{final} = 0.227 + 0.227 + 0.417 + 0.417 = 1.288~\mathrm{atm}$$ The initial and equilibrium total pressures are equal, but this is not true for all gaseous systems. It is only the case when the total number of moles of gas remains constant, as in this particular reaction where the stoichiometric coefficients are equal on both sides of the equilibrium equation.)